Solve first-order recurrences and use initial conditions in second-order models.
Loading your saved responses…
Try
A process starting at 1 and doubling its previous value then adding 3 gives 1,5,13,29,… . The recurrence tells how to continue; a closed formula tells how to reach an index directly. Neither is complete without a starting value.
Predict
Predict how an affine recurrence can become geometric. For aₙ=2aₙ₋₁+3, define bₙ=aₙ+3. Then bₙ=2bₙ₋₁. With a₀=1, b₀=4, so aₙ=4·2ⁿ−3. The shifted sequence exposes the repeated multiplication.
Experiment
Compute the first four terms from the recurrence and independently from the formula. Then change a₀ while retaining the same rule. Observe that the recurrence describes a family of sequences until an initial condition selects one.
Understand
Worked second-order method: for aₙ=3aₙ₋₁−2aₙ₋₂, trying aₙ=rⁿ gives r²−3r+2=0, with roots 1 and 2. A combination A+B2ⁿ satisfies the recurrence. The values a₀=0,a₁=1 give A=−1,B=1. Verify both initial cases and the recurrence to establish aₙ=2ⁿ−1 by induction.
Characteristic roots are a method for finding candidates, not permission to skip verification. Repeated roots need terms such as nrⁿ; nonhomogeneous rules need a particular solution as well. This lesson establishes the basic distinct-root case, not all recurrence theory.
Invalid proof: a formula agrees with the first ten terms, therefore it always agrees. A general substitution and induction step must connect every later index to established ones.
Apply
Keep the index origin and number of required initial values explicit. When proving a second-order formula, show both base cases and explain why the recurrence determines each subsequent value uniquely.
Continue
Depth: divide-and-conquer running times also obey recurrences, but their input domains and rounding matter. A rule on powers of two is not automatically a formula for every integer size. Generating functions provide another way to encode and solve recurrences.
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
a₀=2 and aₙ=3aₙ₋₁+1 for n≥1. Find a₃.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
F₀=0,F₁=1 and Fₙ=Fₙ₋₁+Fₙ₋₂ for n≥2. Find F₇.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
A second-order recurrence normally needs:
A guessed formula solves a recurrence only after checking:
Prove the solution of a₀=1, aₙ=2aₙ₋₁+3 is aₙ=4·2ⁿ−3 for n≥0.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Solve a₀=0,a₁=1,aₙ=3aₙ₋₁−2aₙ₋₂ and prove your result.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
